Metacentre

Metacentre is defined as the point about which a body starts oscillating when a body is tilted by a small angle. 

In other words, the metacentre is defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

Metacentric height:

The distance between the centre of gravity of a floating body and the metacentre is called metacentric height.

•  For stable equilibrium, the position of metacentre M remains higher than c.g. of the body, G.

•  For unstable equilibrium, the position of metacentre M remains lower than G.

•  For neutral equilibrium, the position of metacentre M coincides with G.

Metacentre is defined as the point about which a body starts oscillating when a body is tilted by a small angle. 

In other words, the metacentre is defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

Metacentric height:

The distance between the centre of gravity of a floating body and the metacentre is called metacentric height.

•  For stable equilibrium, the position of metacentre M remains higher than c.g. of the body, G.

•  For unstable equilibrium, the position of metacentre M remains lower than G.

•  For neutral equilibrium, the position of metacentre M coincides with G.

Let us consider a floating body in equilibrium as shown below. Let the position of the centre of gravity and centre of buoyancy be G and B. Let a small angular displacement is given in clockwise direction such that the centre of  position is shifted to a new position `B_1`. The vertical line through `B_1` cuts the normal axis at point M. Thus the point M represent the metacentre and GM should be the corresponding metacentric height.

To find the metacentric height GM, let us  consider an elementary cylindrical prism QQ� of portion N�ON at a distance x from O. Let the area of this elementary prism be dA. The height of this elementary prism is given by x.θ. The volume of this elementary prism is given by:

`dV=x*theta*dA`

The upward force or buoyancy force acting at this prism ` (dP_B)` is given by:

 `dP_B = gamma*dV=gamma*x*theta*dA`

(where, `gamma` = unit weight of liquid)

The moment of this elemental buoyancy force about O is,

`=x*dP_B`

`=gamma*x*theta*dA*x=gamma*theta*dA*x^2`

The total moment is the integration of the elemental moment as,

`M=gamma*theta*int (x^2 dA)`

`M=gamma*theta*I_(y y)`.. (i)

Where ` I_(y y)`  is the moment of inertia of the sectional area at the water line about axis through O.

Now,

As the body is tilted from its original position, it's centre of buoyancy is shifted from B to `B_1` with a distance r. The buoyant force will create a moment with respect to the rolling axis or point O which is equal to,

`M=F_B*r`

From figure,

`r=BM*sintheta`

For small angle,

`sintheta~=theta`

Thus,

`M=F_B*BM*theta`

`M=gamma*V*BM*theta`... (ii)

Where, `V=` volume displaced 

Now from equation (i) and (ii), we get,

`M=gamma*V*BM*theta=gamma*theta*I_(y y)`

or, `BM=I_(y y)/V`

Now from figure,

`GM=BM-BG`

`GM=I_(y y)/V-BG`

This is the required expression for the metacentric height.